Question

For each operation $\ast$ defined below, determine whether $\ast$ is binary, commutative or associative.
(i) On $Z$, define $a\ast b=a-b$
(ii) On $Q$, defined $a\ast b=ab+1$
(iii) On $Q$, defined $a\ast b=\frac{ab}{2}$
(iv) On $Z^{+}$, defined $a\ast b=2^{ab}$
(v) On $Z^{+}$, defined $a\ast b=a^b$
(vi) On $R-\{-1\}$, defined $a\ast b=\frac{a}{b+1}$

Answer 1

i)On $z$, define $a\ast b=a-b$

here $aϵz^{+}$ and $bϵz^{+}$

i.e.,$a$ and $b$ are positive integers

Let $a=2,b=5\Rightarrow 2\ast 5=2-5=-3$

But $-3$ is not a positive integer

i.e., $-3\notin z^{+}$

hence,$\ast$ is not a binary operation.

ii)On $Q$,define $a\ast b=ab-1$

Check commutative

$\ast$ is commutative if,$a\ast b=b\ast a$

$a\ast b=ab+1;a\ast b=ab+1=ab+1$

Since $a\ast b=b\ast aforalla,bϵQ$

$\ast$ is commutative.

Check associative

$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$

$(a\ast b)\ast c=(ab+1)\ast c=(ab+1)c+1=abc+c+1a\ast (b\ast c)=a\ast (bc+1)=a(bc+1)+1=abc+a+1$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

iii)On $Q$,define $a\ast b=\frac{ab}{2}$

Check commutative

$\ast$ is commutative is $a\ast b=b\ast a$

$a\ast b=\frac{ab}{2}b\ast a=\frac{ba}{2}=\frac{ab}{2}a\ast b=b\ast a\forall a,bϵQ$

$\ast$ is commutativve.

Check associative

$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$

$(a\ast b)\ast c=\frac{\left(\frac{ab}{2}\right)\ast c}{2}=\frac{abc}{4}$

$(a\ast b)\ast c=a\ast (b\ast c)=\frac{a\times \frac{bc}{2}}{2}=\frac{abc}{4}$

Since $(a\ast b)\ast c=a\ast (b\ast c)\forall a,b,cϵQ$

$\ast$ is an associative binary operation.

iv)On $z^{+}$, define if $a\ast b=b\ast a$

$a\ast b=2^{ab}b\ast a=2^{ba}=2^{ab}$

Since $a\ast b=b\ast a\forall a,b,cϵz^{+}$

$\ast$ is commutative.

Check associative.

$\ast$ is associative if $

$(a\ast b)\ast c=a\ast (b\ast c)(a\ast b)\ast c={\left(2^{ab}\right)}^{\ast }c=2^{2^{ab}}ca\ast (b\ast c)=a\ast \left(2^{ab}\right)=2^{a{2}^{bc}}$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

v)On $z^{+}$ define $a\ast b=a^b$

$a\ast b=a^b,b\ast a=b^a\Rightarrow a\ast b\ne b\ast a$

$\ast$ is not commutative.

Check associative

$\ast$ is associative if $

$(a\ast b)\ast c=a\ast (b\ast c)(a\ast b)\ast c={\left(a^b\right)}^{\ast }c={\left(a^b\right)}^{c}a\ast (b\ast c)=a\ast \left(2^{bc}\right)=2^{a{2}^{bc}}$

$eg:-Leta=2,b=3$ and $c=4$

$(a\ast b{)}^{\ast }c=(2\ast 3{)}^{\ast }4=(2^3{)}^{\ast }4=8\ast 4=8^{4}a\ast (b\ast c)=2^{\ast }(3\ast 4)=2^{\ast }(3^4)=2\ast 81=2^{81}$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not an associative binary operation.

vi)On $R-\{-1\},$ define $a\ast b=\frac{a}{b+1}$

Check commutative

$\ast$ is commutative if $a\ast b=b\ast a$

$a\ast b=\frac{a}{b+1}b\ast a=\frac{b}{a+1}$

Since $a\ast b\ne b\ast a$

$\ast$ is not commutatie.

Check associative

$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$

$(a\ast b)\ast c=(\frac{a}{b+1}{)}^{\ast }c=\frac{\frac{a}{b}+1}{c}=\frac{a}{c(b+1)}a\ast (b\ast c)=a\ast (\frac{b}{c+1})=\frac{\frac{a}{b}}{c+1}=\frac{a(c+1)}{b}$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not a associative binary operation.